HDU 2890 Longest Repeated subsequence

HDU_2890

    在罗穗骞的论文里有说过至少重复K次的可重叠的最长子串该怎么找，只不过这个题改成了不可重叠。我用的是比较暴力的办法，把按height值分组之后每组内sa[i]的值提取出来排个序，然后扫一遍看能不能形成至少K个不重叠的子串。

    此外这个题有两个大坑：①题目中所谓的subsequence应该是substring，在HIT的OJ上这个题目的描述就都是用的substring；②所谓的字典序可能会带来歧义，比如结果有两种可能5 100和5 99，如果将整数当成字符串来处理显然是5 100应该字典序更小，因为100的字典序比99小，但就这个题目而言应该是5 99的字典序比较小，也就是说要把每个整数当成一个字符来看待。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define MAXD 50010
char a[MAXD];
int N, K, q[MAXD], FLAG;
struct El
{
    int id, v, r;
}el[MAXD];
bool cmp1(const El &x, const El &y)
{
    return x.v < y.v;
}
bool cmp2(const El &x, const El &y)
{
    return x.id < y.id;
}
struct Suffix
{
    int r[MAXD], rank[MAXD], sa[MAXD], height[MAXD], ws[MAXD], wv[MAXD], wa[MAXD], wb[MAXD];
    int cmp(int *p, int x, int y, int l)
    {
        return p[x] == p[y] && p[x + l] == p[y + l];
    }
    void da(int n, int m)
    {
        int i, j, p, *x = wa, *y = wb, *t;
        for(i = 0; i < m; i ++) ws[i] = 0;
        for(i = 0; i < n; i ++) ++ ws[x[i] = r[i]];
        for(i = 1; i < m; i ++) ws[i] += ws[i - 1];
        for(i = n - 1; i >= 0; i --) sa[-- ws[x[i]]] = i;
        for(j = p = 1; p < n; j *= 2, m = p)
        {
            for(p = 0, i = n - j; i < n; i ++)
                y[p ++] = i;
            for(i = 0; i < n; i ++)
                if(sa[i] >= j)
                    y[p ++] = sa[i] - j;
               for(i = 0; i < n; i ++) wv[i] = x[y[i]];
            for(i = 0; i < m; i ++) ws[i] = 0;
            for(i = 0; i < n; i ++) ++ ws[wv[i]];
            for(i = 1; i < m; i ++) ws[i] += ws[i - 1];
            for(i = n - 1; i >= 0; i --) sa[-- ws[wv[i]]] = y[i];
            for(t = x, x = y, y = t, x[sa[0]] = 0, i = p = 1; i < n; i ++)
                x[sa[i]] = cmp(y, sa[i - 1], sa[i], j) ? p - 1 : p ++;
           }
    }
    void calheight(int n)
    {
        int i, j, k = 0;
           for(i = 1; i <= n; i ++)
        rank[sa[i]] = i;
        for(i = 0; i < n; height[rank[i ++]] = k)
            for(k ? -- k : 0, j = sa[rank[i] - 1]; r[i + k] == r[j + k]; k ++);
    }
    void init()
    {
        int i, cnt;
        for(i = 0; i < N; i ++)
        {
            scanf("%d", &el[i].v);
            el[i].id = i;
        }
        std::sort(el, el + N, cmp1);
        el[0].r = cnt = 1;
        for(i = 1; i < N; i ++) el[i].r = el[i].v == el[i - 1].v ? cnt : ++ cnt;
        std::sort(el, el + N, cmp2);
        for(i = 0; i < N; i ++) r[i] = el[i].r;
        r[N] = 0;
        da(N + 1, N + 1);
        calheight(N);
    }
    int check(int k, int rear)
    {
        int i, pre, cnt;
        std::sort(q, q + rear);
        pre = q[0], cnt = 1;
        for(i = 1; i < rear; i ++) if(q[i] - pre >= k) ++ cnt, pre = q[i];
        return cnt >= K;
    }
    int deal(int k)
    {
        int i, j, rear = 0;
        for(i = 1; i <= N; i ++)
        {
            if(height[i] < k)
            {
                if(check(k, rear))
                {
                    FLAG = sa[i - 1];
                    return 1;    
                }
                q[0] = sa[i], rear = 1;
            }
            else q[rear ++] = sa[i];
        }
        if(check(k, rear))
        {
            FLAG = sa[i - 1];
            return 1;    
        }
        return 0;
    }
    void solve()
    {
        int i, j, min = 1, mid, max = N / 2 + 1;
        init();
        for(;;)
        {
            mid = min + max >> 1;
            if(min == mid) break;
            if(deal(mid)) min = mid;
            else max = mid;
        }
        deal(mid);
        printf("%d\n", mid);
        for(i = FLAG, j = 0; j < mid; i ++, j ++) printf("%d\n", el[i].v);
    }
}suffix;
void solve()
{
    scanf("%d%d", &N, &K);
    suffix.solve();
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t --)
    {
        solve();
        if(t) printf("\n");
    }
    return 0;
}